Which of the following numbers is a multiple of 3? ${52,53,58,103,108}$
The multiples of $3$ are $3$ $6$ $9$ $12$ ..... In general, any number that leaves no remainder when divided by $3$ is considered a multiple of $3$ We can start by dividing each of our answer choices by $3$ $52 \div 3 = 17\text{ R }1$ $53 \div 3 = 17\text{ R }2$ $58 \div 3 = 19\text{ R }1$ $103 \div 3 = 34\text{ R }1$ $108 \div 3 = 36$ The only answer choice that leaves no remainder after the division is $108$ $ 36$ $3$ $108$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $3$ are contained within the prime factors of $108$ $108 = 2\times2\times3\times3\times3 3 = 3$ Therefore the only multiple of $3$ out of our choices is $108$. We can say that $108$ is divisible by $3$.